Question

A point $P$ is given on the circumference of a circle of radius $r$. The chord $QR$ is parallel to the tangent line at $P$. The maximum area of the triangle $PQR$ is:

• A. $\frac{3\sqrt{2}}{4}{r}^2$
• B. $\frac{3\sqrt{3}}{4}{r}^2$
• C. $\frac{3}{8}r$
• D. $\frac{3\sqrt{2}}{4}r$

Answer 1

Answer: (B)

$\frac{3\sqrt{3}}{4}{r}^2$

Essentially, since the tangent at $P$ is parallel to the chord $QR$, the perpendicular bisector $PB$ passes through the center $C$.

Now, $\Delta PBQ$ is congruent to $\Delta PBR$.

$PB$ common, $QB=BR$, and $(\angle PBQ=\angle PBR=90$ degree)

So we just need to maximize the area of one of the triangles:

$\frac{1}{2}\times (r+r\cos \theta )\times r\sin \theta$, ( where $\theta =\angle BCQ,PB=r+r\cos \theta ,BQ=r\sin \theta )$

We do the usual derivative based maximization:

$\frac{d}{d\theta }[\frac{1}{2}\times (r+r\cos \theta )\times r\sin \theta ]=0$

Solving, we get

$\frac{1}{2}[r^2\cos \theta +r^2\times ({\cos }^2\theta -{\sin }^2\theta \left)\right]=0$

$\Rightarrow 2\times {\cos }^2\theta +\cos \theta -1=0$

Hence, $\theta =0$ or $60$ degree .(obviously we throw 0 out, it gives the global minima).

So, $\theta =60$ degree and the maximum area of $\Delta PQR$ is $\left[\right(r+r\cos 60)\times r\sin 60]$, which is $\frac{3}{4}\times \sqrt{3}\times r^2$